POJ2185 Milking Grid

Milking Grid

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 8248		Accepted: 3557

Description

Every morning when they are milked, the Farmer John's cows form a rectangular grid that is R (1 <= R <= 10,000) rows by C (1 <= C <= 75) columns. As we all know, Farmer John is quite the expert on cow behavior, and is currently writing a book about feeding behavior in cows. He notices that if each cow is labeled with an uppercase letter indicating its breed, the two-dimensional pattern formed by his cows during milking sometimes seems to be made from smaller repeating rectangular patterns.

Help FJ find the rectangular unit of smallest area that can be repetitively tiled to make up the entire milking grid. Note that the dimensions of the small rectangular unit do not necessarily need to divide evenly the dimensions of the entire milking grid, as indicated in the sample input below.

Input

* Line 1: Two space-separated integers: R and C

* Lines 2..R+1: The grid that the cows form, with an uppercase letter denoting each cow's breed. Each of the R input lines has C characters with no space or other intervening character.

Output

* Line 1: The area of the smallest unit from which the grid is formed

Sample Input

2 5
ABABA
ABABA

Sample Output

Hint

The entire milking grid can be constructed from repetitions of the pattern 'AB'.

Source

USACO 2003 Fall

题意：

　　在N*M字符矩阵中找出一个最小子矩阵，使其多次复制所得的矩阵包含原矩阵。N<=10000,M<=75

做法：

　　先找出最大的K，使得原矩阵是若干个K*M的矩阵拼成一列后的子矩阵

　　把一行看做一个整体，对列做KMP

　　用应用1的方法确定最小行宽

　　再在K*M的矩阵中，把一列看做一个整体，用同样的方法求最小行宽

　　O(N*M)

#include<cstdio>
#include<cstring>
using namespace std;
const int N=1e4+5;
const int M=80;
char s[N][M];
int n,m;
int l1,repetend1,fail1[N];
int l2,repetend2,fail2[M];
long long h1[N],h2[M];
void get_next1(){
    int p=0;fail1[1]=0;
    for(int i=2;i<=l1;i++){
        while(p>0&&h1[i]!=h1[p+1]) p=fail1[p];
        if(h1[i]==h1[p+1]) p++;
        fail1[i]=p;
    }
    repetend1=l1-fail1[l1];
}
void get_next2(){
    int p=0;fail2[1]=0;
    for(int i=2;i<=l2;i++){
        while(p>0&&h2[i]!=h2[p+1]) p=fail2[p];
        if(h2[i]==h2[p+1]) p++;
        fail2[i]=p;
    }
    repetend2=l2-fail2[l2];
}
#define P 29
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++) scanf("%s",s[i]+1);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            h1[i]=h1[i]*P+s[i][j];
        }
    }
    for(int i=1;i<=m;i++){
        for(int j=1;j<=n;j++){
            h2[i]=h2[i]*P+s[j][i];
        }
    }
    l1=n;get_next1();
    l2=m;get_next2();
    printf("%d",repetend1*repetend2);
    return 0;
}

__EOF__

本文作者：shenben
本文链接：https://www.cnblogs.com/shenben/p/6382910.html
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